The discussion revolves around the confusion in vector addition, particularly when dealing with forces represented at angles of 360 and 270 degrees. Participants are examining how to correctly interpret the components of these forces in a coordinate system.
Multiple interpretations of the problems are being explored, with some participants suggesting that the original solutions may contain errors. Guidance is offered regarding the treatment of tension and compression forces, as well as the need to consistently apply signs to the components based on their directions.
There is an ongoing debate about the correctness of the provided solutions, particularly regarding the treatment of angles and the resultant forces. Some participants express uncertainty about the assumptions made in the original problems, particularly concerning the definitions of tension and compression.
uniquesoljuh said:When adding the resultant of two or more forces i get confused when adding forces that are on 360 and 270 degrees. Like the question is 6 Newtons on 360 degrees, i don't get how they get -6, when cos 360 is 1
can anyone help me??
uniquesoljuh said:The original probem asks to calculate the resultant of the three forces
F1 = 6kN at (0 or 360 degrees)
F2 = 8 Kn at 240 degrees
F3 = 10 Kn at 210 degrees
The solution they show is:-
Fx= 8cos240-6-10cos210
I understand everything except how they get -6. And by the way all the forces are in tension(heading outwards)
uniquesoljuh said:Yea i agree with you but they can't do the same error twice. This other questions has 4 forces:
F1 = 60 at 40 degrees (tension)
F2 = 50 at 110 degrees (compression)
F3 = 30 at 160 degrees (tension)
F4 = 40 at 270 degrees (compression)
when adding the Fy components they have F4 as positive when it should be negative because sine 270 is negative. Do you think this question is all incorrect?
uniquesoljuh said:The answers are
Rx = 34.87 Kn
Ry = 41.84 Kn
R = 54.5 Kn
angle = 50.2
uniquesoljuh said:Sorry for any inconvenience guys, but the answers given are correct because i asked my cousin and he said "hypythetically, if you have 3 Fy forces are going in one direction e.g. all facing down (which are negative) you can make them positive and make the other force negative. In other words you have to consistent with you the signs like make all forces in one direction positive or negative and vice verse to the other forces
uniquesoljuh said:Yea i agree with you but they can't do the same error twice. This other questions has 4 forces:
F1 = 60 at 40 degrees (tension)
F2 = 50 at 110 degrees (compression)
F3 = 30 at 160 degrees (tension)
F4 = 40 at 270 degrees (compression)
when adding the Fy components they have F4 as positive when it should be negative because sine 270 is negative. Do you think this question is all incorrect?
<br /> <br /> <br /> <br /> Thanks for the reply i appreciate the example you gave. But i thought when it comes to vectors it doesn't matter if they're in tension or compression unless its a truss because we only want the magnitude of each force. F2 and F4 were both in compression however F2 was negative but F4 was positivezgozvrm said:Wrong! F4 should not be negative "because sin(270) is negative." It is negative because it is in compression (i.e. it points inward, rather than outward).
F2 should also be subtracted because it is in compression (even though sin(110) is positive).
You should think of these forces as pulling or pushing on a single point. Think of that point as the origin. You then add all the forces that are pulling the point (tension) and subtract all forced that are pushing on it (compression).
Consider a simpler scenario: Two forces pulling on a common point in opposite directions:
F1 = 52n at 300 deg (or, more correctly -60 deg)
F2 = 37n at 120 deg
The resultant force, FR, is the vector sum of F1 and F2, that is F1 + F2
To do this, we generally need to break down each vector into its x- and y- components, add the corresponding components together, and calculate the new force give the new components.
So (leaving off units for simplicity - we can add them back in later),
F1_x = 52 cos(300)[/tex]<br /> F2_x = 37 cos(120)[/tex]<br /> F1_y = 52 sin(300)[/tex]&lt;br /&gt; F2_y = 37 sin(120)[/tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; This gives:&amp;lt;br /&amp;gt; FR_x = F1_x + F2_x = 52 cos(300) + 37 cos(120)[/tex]&amp;amp;lt;br /&amp;amp;gt; FR_y = F1_y + F2_y = 52 sin(300) + 37 sin(120)[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; And finally, the magnitude of FR can be found by:&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; \sqrt{(FR_x)^2 + (FR_y)^2}&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; with direction:&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; tan^{-1} \left( \frac{FR_y}{FR_x} \right)&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Which comes out to FR = 15n at 300 deg (-60 deg)&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; This is what you should have already expected, since the forces are directly opposite each other (180 degrees apart).&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; When 2 forces are pulling in opposite directions, there are 3 possible outcomes for the resultant force:&amp;amp;amp;lt;br /&amp;amp;amp;gt; 1) the resultant force pulls in the same direction as F1 (with a magnitude less than that of F1)&amp;amp;amp;lt;br /&amp;amp;amp;gt; 2) the resultant force pulls in the same direction as F2 (with a magnitude less than that of F2)&amp;amp;amp;lt;br /&amp;amp;amp;gt; 3) there is no resultant force, because the 2 original forces are equal in magnitude, thus canceling each other out.&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Now, if instead we had 2 forces, one of which is pulling on a point and the other that is pushing on it from the opposite side, we essentially have to forces acting on that point in the same direction. Using the forces listed above, let&amp;amp;amp;amp;#039;s assume F1 is pulling on the point, while F2 is pushing. Then, the resultant force becomes FR = F1 - F2 because these forces are acting on a point in different ways. Whether the sine of one angle or another is negative has no bearing on the equation that you use to find the resultant.&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; I think what your cousin may have been trying to point out is that you can look at a &amp;amp;amp;amp;quot;negative force&amp;amp;amp;amp;quot; as a positive force in the opposite direction. That is since we have F1 as a tension force and F2 as a compression force, we can consider F2 to be a &amp;amp;amp;amp;quot;negative force&amp;amp;amp;amp;quot; (in respect to the point being acted upon). Therefore, we can think of F2 as being either (-37)n at 120 deg or as (+37)n at 300 deg. If we do that, &amp;amp;amp;lt;i&amp;amp;amp;gt;then&amp;amp;amp;lt;/i&amp;amp;amp;gt; we can add the forces. This really is the same as adding real numbers: A - B = A + (-B)
uniquesoljuh said:Thanks for the reply i appreciate the example you gave. But i thought when it comes to vectors it doesn't matter if they're in tension or compression unless its a truss because we only want the magnitude of each force. F2 and F4 were both in compression however F2 was negative but F4 was positive
uniquesoljuh said:Mindscrape,
I saw a different question and yea you are right but then let me ask you this; If you have a 10 kN vector that's on 270 degrees and it in compression, will the positive right because -(-9)=9?? is that correct
._|evo|_. said:Request post to be deleted, put in a separate topic please. Thanks.
uniquesoljuh said:Mindscrape,
I saw a different question and yea you are right but then let me ask you this; If you have a 10 kN vector that's on 270 degrees and it in compression, will the positive right because -(-9)=9?? is that correct